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3 years ago

9

A 200g particle oscillating in SHM travels 66cm between the two extreme points in its motion with an average speed of 110cm/s. F

ind: a) The angular frequency. s−1 b) The maximum force on the particle. N b) The maximum speed.

1 answer:

IgorC [24]3 years ago

4 0

For a:v = d / Δt
110 = 0.66 / Δt
Δt = 0.66 / 110
Δt = 0.006 s
the period is:
T = 2Δt
T = 2*0.006
T = 0.012 s
the frequency is the inverse of the period. so: f = 1 / T
f = 83.3333333 Hz (about; Hz = 1/s)
b. T = 2π√(m/k)
being the mass m = 200g = 0.2 kg = 2*10^-1 kg, π = 3.14 (about) and T = 0.012, k is equal to:
0.012 = 6.28√(2*10^-1 / k)
0.012 / 6.28 = √(2*10^-1 / k)
0.00191082803 = √(2*10^-1 / k)
2*10^-1/ k = 0.000003
2*10^-1 / k = 3*10^-6
k = 2*10^-1 / 3*10^-6
k = 6.67*10^-5

now using hooke's law:
F = -kx
F = - 6.67*10^-5* 3.3*10^-1
F = -2.20x10^-5m
F = -0.22 *10^4 N

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